（ヅ） 写了： 2025年 1月 28日 16:15
半个小时我的本地deepseek-r1:32b 终于出完结果嘞
>>> integer solution of x^2 + x + 1 = 0 ( mod 11)
<think>
Alright, so I need to find all integer solutions for the equation \(x^2 + x + 1 = 0 \mod{11}\). Hmm, okay. Let me
think about how to approach this.
First off, solving quadratic equations modulo a prime number like 11 can sometimes be done by trying out all
possible residues from 0 up to 10 and seeing which ones satisfy the equation. Since 11 is not too large, that
might be manageable. But before jumping into that, maybe there's a smarter way.
I remember something about quadratic equations modulo primes. There's this thing called the discriminant. For a
general quadratic equation \(ax^2 + bx + c = 0\), the discriminant is \(D = b^2 - 4ac\). If the discriminant is a
square modulo the prime, then there are solutions; otherwise, there aren't.
So in this case, let's compute the discriminant. Here, \(a = 1\), \(b = 1\), and \(c = 1\), so:
\(D = (1)^2 - 4(1)(1) = 1 - 4 = -3\).
Hmm, okay, so we have a discriminant of \(-3\) modulo 11. I need to check if \(-3\) is a quadratic residue modulo
11. In other words, does there exist some number \(y\) such that \(y^2 \equiv -3 \mod{11}\)?
Let me compute \(-3\) mod 11 first. Since 11 - 3 = 8, so \(-3 \equiv 8 \mod{11}\). So I need to check if 8 is a
quadratic residue modulo 11.
To do that, one method is to use Euler's criterion, which says that for an odd prime \(p\), and integer \(a\) not
divisible by \(p\), \(a^{(p-1)/2} \equiv 1 \mod{p}\) if \(a\) is a quadratic residue, and \(-1\) otherwise.
So let me compute \(8^{5} \mod{11}\). Why 5? Because (11 - 1)/2 = 5. Let's calculate:
\(8^1 = 8\)
\(8^2 = 64\), which is 64 - 5*11=64-55=9 mod 11.
\(8^3 = 8 * 9 = 72\). 72 divided by 11: 6*11=66, so 72-66=6. So \(8^3 \equiv 6 \mod{11}\).
\(8^4 = 8 * 6 = 48\), which is 48 - 4*11=48-44=4 mod 11.
\(8^5 = 8 * 4 = 32\). 32 - 2*11=32-22=10 mod 11. So \(8^5 \equiv 10 \mod{11}\).
Since 10 is not congruent to 1, Euler's criterion tells me that 8 is not a quadratic residue modulo 11. Therefore,
the discriminant isn't a square in this field, which suggests that there are no solutions.
Wait, but let me double-check because sometimes even if I think I'm correct, it's better to be thorough. Maybe
trying all possible residues from 0 to 10 will confirm whether any satisfy the equation \(x^2 + x + 1 \equiv 0
\mod{11}\).
Alright, let me create a table of values for each x in [0, 10] and compute \(x^2 + x + 1\) mod 11.
Starting with x=0:
\(0^2 + 0 + 1 = 1\). 1 mod 11 is 1. Not zero.
x=1:
\(1 + 1 + 1 = 3\). 3 mod 11 is 3. Nope.
x=2:
4 + 2 + 1 =7. 7 mod 11 is 7. Still not.
x=3:
9 + 3 + 1=13. 13-11=2. Not zero.
x=4:
16 + 4 + 1=21. 21 - 2*11=21-22=-1≡10 mod11. Still not zero.
x=5:
25 +5 +1=31. 31 -2*11=9. Not zero.
x=6:
36 +6 +1=43. 43 - 3*11=43-33=10. Not zero.
x=7:
49 +7 +1=57. 57 - 5*11=57-55=2. Nope.
x=8:
64 +8 +1=73. 73 -6*11=73-66=7. Not zero.
x=9:
81 +9 +1=91. 91 -8*11=91-88=3. Nope.
x=10:
100 +10 +1=111. 111 -10*11=111-110=1. Not zero.
Hmm, so none of the residues from 0 to 10 satisfy the equation \(x^2 + x + 1 \equiv 0 \mod{11}\). So that confirms
it: there are no solutions modulo 11 for this quadratic equation.
Wait a second, but I just remembered something. Quadratic equations over finite fields can have 0, 1, or 2
solutions. Since the discriminant wasn't a square, we have zero solutions here, which matches with our exhaustive
check above.
Alternatively, maybe there's another method involving completing the square? Let me try that for practice.
Starting with \(x^2 + x + 1 \equiv 0 \mod{11}\).
Completing the square:
\(x^2 + x = -1 \mod{11}\)
Take half of the coefficient of x, which is 1/2. Since we're working mod11, 1/2 is equivalent to 6 because
2*6=12≡1 mod11.
So, \( (x + 6)^2 - 36 ≡ -1 \mod{11}\).
Wait, let me compute that step by step:
\(x^2 + x = (x + 1/2)^2 - (1/2)^2\). So,
\( (x + 6)^2 - 36 ≡ -1 \mod{11} \)
Compute \( -36 mod11 \): 36 /11 is 3*11=33, so 36-33=3. Thus, -36≡-3≡8 mod11.
So:
\( (x + 6)^2 ≡ -1 + 36 mod11\), which simplifies to \( (x + 6)^2 ≡ (-1) + (36 mod11) \).
Wait, hold on. Let me redo that step because I might have messed up.
We had:
\( x^2 + x = -1 \mod{11} \)
Complete the square:
\( x^2 + x + (1/4) = -1 + (1/4) \mod{11} \)
So,
\( (x + 1/2)^2 ≡ (-1) + (1/4) mod11 \)
Compute \(1/4\) modulo11: since 4*3=12≡1 mod11, so 1/4≡3 mod11.
Thus:
Left side: \( (x + 6)^2 \mod{11} \) because 1/2 is 6 in mod11.
Right side: (-1) + 3 ≡ 2 mod11.
So the equation becomes:
\( (x + 6)^2 ≡ 2 mod11 \)
Now, so we need to solve \( y^2 ≡ 2 mod11 \), where y = x +6. So does 2 have square roots modulo11?
Let me check which squares give 2 mod11.
Compute squares of numbers from 0 to 10:
0²=0
1²=1
2²=4
3²=9
4²=16≡5
5²=25≡3
6²=36≡3
7²=49≡5
8²=64≡9
9²=81≡4
10²=100≡1.
So the squares modulo11 are 0,1,3,4,5,9. So 2 isn't a square modulo11 either. Therefore, equation \( y^2 ≡2
\mod{11}\) has no solution, so original equation also has no solutions.
Wait, but earlier when computing Euler's criterion for discriminant 8 (from the first method), we saw that 8 was
not a quadratic residue as well? Or wait, in the second method, we arrived at needing to solve y²≡2. So yeah, same
result: no solution.
So all methods converge on there being no solutions to this equation in integers mod11.
Therefore, I think it's settled. There are no integer solutions modulo11 for \( x^2 +x +1 ≡0 \mod{11} \).
**Final Answer**
There are no integer solutions. The final answer is \boxed{\text{No solution}}.
</think>
To determine the integer solutions of the quadratic equation \(x^2 + x + 1 \equiv 0 \pmod{11}\), we can use
several methods:
1. **Exhaustive Check**:
- We check all residues from 0 to 10 modulo 11.
- None of these values satisfy the equation \(x^2 + x + 1 \equiv 0 \pmod{11}\).
2. **Quadratic Residue Check**:
- The discriminant of the quadratic equation is calculated as \(\Delta = b^2 - 4ac = 1 - 4 = -3\).
- Converting \(-3\) modulo 11, we get \(8\). We check if 8 is a quadratic residue modulo 11.
- The squares modulo 11 are calculated: 0, 1, 3, 4, 5, 9. Since 8 is not among these, it is not a quadratic
residue.
3. **Completing the Square**:
- Rewrite the equation as \(x^2 + x = -1 \pmod{11}\).
- Complete the square: \((x + 6)^2 - 36 \equiv -1 \pmod{11}\).
- Simplifying, we get \((x + 6)^2 \equiv 2 \pmod{11}\). Checking if 2 is a quadratic residue modulo 11, we find
it is not.
Since all methods confirm that there are no solutions, we conclude:
\[
\boxed{\text{No solution}}
\]
